Posted by Robert on Wednesday, July 25, 2012 at 2:17pm.
This is the last question for a final exam preparation sheet and I can't figure out the answer. Any help would be greatly appreciated. Thank you.
 One very long wire carries current 10.0 A to the left along the x axis. A second very long wire carries current 75.0 A to the right along the line
(y = 0.280 m, z = 0).
(a) Where in the plane of the two wires is the total magnetic field equal to zero? (along the y axis)
(b) A particle with a charge of 2.00 µC is moving with a velocity of 150 Mm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle.
(c) What If? A uniform electric field is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field.
 Physics  Electromagnetism  Elena, Wednesday, July 25, 2012 at 4:39pm
(a)
Magnetic field of the long current carrying wire according to BiotSavert Law is
B=μₒI/2•π•a
The desired point is separated by distance “x” from the 1st current and by distance (d+x) from the 2nd current (along yaxis).
μₒI1/2•π•x=μₒI2/2•π•(d+x),
10•(0.28+x)=75•x
2.8+10x=75 x.
65x=2.8
x=0.431 m.
(b) Magnetic force (Lorentz force) is
F=qvB sinα.
Since the motion of the particle is along the straight line, then F=0
(c) Electric field has to be applied along the direction of the particle motion (E↑↑ῡ, E↑↓ῡ)
 Physics  Electromagnetism  Robert, Thursday, July 26, 2012 at 9:09am
thank you Elena!
 Physics  Electromagnetism  Robin, Wednesday, June 29, 2016 at 6:10pm
For people of the future, Elena's answer for part B is wrong. It's better to use the equation
F = q*v(cross)B
where B is the sum of the B field created by the 50 A wire and the 30 A wire.
> 50A

 .18m

> v = 150E6 m/s

 .1m
< 30A
B_tot = mu/2pi * (50/.18 (k) + 30/.1 (k))
= 1.2E4 (k) Teslas
Then you take the cross product and eventually get F = .035j N
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